OSCILLATIONS MCQs for NEET UG PHYSICS — Medium

1 Medium

Application +4 | -1

A mass m is suspended from a spring of length l and force constant K. The frequency of vibration of the mass is

{f_1}

. The spring is cut into two equal parts and the same mass is suspended from one of the parts. The new frequency of vibration of mass is

{f_2}

. Which of the following relations between the frequencies is correct

View answer and explanation

Correct option: D

{f_2} = \sqrt 2 {f_1}

Option D (

{f_2} = \sqrt 2 {f_1}

) is correct because cutting the spring in half doubles the spring constant, which increases the frequency by a factor of

\sqrt{2}

.

Solution approach

Correct Answer + Core Reason: ${f_2} = \sqrt 2 {f_1}$ . Cutting the spring in half doubles the spring constant.

Step-by-Step Solution:

Step 1: The frequency of vibration of a mass-spring system is given by $f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$ , where $k$ is the spring constant and $m$ is the mass.

Step 2: When the spring is cut into two equal parts, the spring constant of each part becomes $2k$ . This is because $k \propto \frac{1}{l}$ , where $l$ is the length of the spring. So, if the length is halved, the spring constant is doubled.

Step 3: The new frequency $f_2$ is given by $f_2 = \frac{1}{2\pi} \sqrt{\frac{2k}{m}}$ .

Step 4: Divide the new frequency by the original frequency: $\frac{f_2}{f_1} = \frac{\frac{1}{2\pi} \sqrt{\frac{2k}{m}}}{\frac{1}{2\pi} \sqrt{\frac{k}{m}}} = \sqrt{\frac{2k}{m} \times \frac{m}{k}} = \sqrt{2}$ .

Step 5: Therefore, $f_2 = \sqrt{2} f_1$ .

Why other options are wrong:
- Option A: Incorrect. The new frequency is $\sqrt{2}$ times the original frequency, not $\frac{1}{\sqrt{2}}$ times.
- Option B: Incorrect. The frequency changes because the spring constant changes.
- Option C: Incorrect. The new frequency is $\sqrt{2}$ times the original frequency, not 2 times.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Frequency of oscillation of a spring-mass system: $f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$
- Spring constant and length relationship: $k \propto \frac{1}{l}$

Memory Aids:
- Remember that cutting a spring in half doubles its spring constant.

Quick Tips:
- When a spring is cut into $n$ equal parts, the spring constant of each part becomes $nk$ .

2 Medium

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Understanding +4 | -1

The time period of a loaded spring on earth is 6s. On the surface of moon, the time period of the same loaded spring will be.

View answer and explanation

Correct option: D

6 s

Option D (6 s) is correct because the time period of a spring-mass system depends on the mass and spring constant, which remain the same on the Moon.

Solution approach

Correct Answer + Core Reason: 6 s. The time period of a spring-mass system is independent of gravity.

Step-by-Step Solution:

Step 1: Recall the formula for the time period of a spring-mass system: $T = 2\pi \sqrt{\frac{m}{k}}$ , where $m$ is the mass and $k$ is the spring constant.

Step 2: The time period depends only on the mass and the spring constant, which are the same on Earth and the Moon.

Step 3: Therefore, the time period remains the same, which is 6 s.

Why other options are wrong:
- Option A: $\sqrt 6 \,s$ : This is incorrect because the time period is independent of gravity.
- Option B: $6\sqrt 6 \,s$ : This is incorrect because the time period is independent of gravity.
- Option C: 1 s: This is incorrect because the time period is independent of gravity.

Key Concepts:
- Spring-mass system
- Time period of a spring-mass system

Memory Aids:
- The time period of a spring-mass system depends only on the mass and spring constant.

Quick Tips:
- Remember that the time period is independent of gravity.

3 Medium

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Analyzing +4 | -1

S-I: A particle executes SHM with frequency

f

. The frequency with which total mechanical energy of particle oscillates is

2 f

.
S-II: Both kinetic as well as potential energy of particle in SHM oscillate with a frequency

2 f.

View answer and explanation

Correct option: D

S-I is false

\&

S-II is true

Option D (S-I is false & S-II is true) is correct because the frequency of oscillation of total mechanical energy is constant, while kinetic and potential energies oscillate at twice the frequency of SHM.

Solution approach

Correct Answer + Core Reason: S-I is false $\& S-II is true. The total mechanical energy is constant, while kinetic and potential energies oscillate at twice the frequency of SHM.\n\n**Step-by-Step Solution:**\n\n**Step 1:** Statement I: A particle executes SHM with frequency$ f $. The frequency with which total mechanical energy of particle oscillates is$ 2f $. This is false. The total mechanical energy in SHM is constant, so it doesn't oscillate. Therefore, its frequency of oscillation is 0.\n\n**Step 2:** Statement II: Both kinetic as well as potential energy of particle in SHM oscillate with a frequency$ 2f $. This is true. The kinetic and potential energies vary sinusoidally with time, and their frequency is twice the frequency of the SHM.\n\n**Why other options are wrong:**\n- **Option A:** S-I$ \& S-II are true. This is incorrect because Statement I is false.
- Option B: S-I $\& S-II are false. This is incorrect because Statement II is true.\n- **Option C:** S-I is true$ \& S-II is false. This is incorrect because Statement I is false and Statement II is true.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Kinetic Energy in SHM
- Potential Energy in SHM
- Total Mechanical Energy in SHM

Memory Aids:
- Remember that the total mechanical energy in SHM is constant.
- Remember that the kinetic and potential energies oscillate with twice the frequency of SHM.

Quick Tips:
- Understand the energy variations in SHM.
- Know that total energy is constant, while kinetic and potential energies oscillate.

4 Medium

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Application +4 | -1

A particle of mass m is executing oscillations about the origin on the x-axis. Its potential energy is

U\left( x \right) = k{\left[ x \right]^3}

, where k is a positive constant. If the amplitude of oscillation is a, then its time period T is

View answer and explanation

Correct option: A

Proportional to

\frac{1}{{\sqrt a }}

Option A (Proportional to

\frac{1}{{\sqrt a }}

) is correct because the time period of oscillation is inversely proportional to the square root of the amplitude when the potential energy is proportional to

x^3

.

Solution approach

Correct Answer + Core Reason: Proportional to $\frac{1}{{\sqrt a }}$ . The time period is inversely proportional to the square root of the amplitude.

Step-by-Step Solution:

Step 1: Given potential energy $U(x) = k[x]

5 Medium

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Analyzing +4 | -1

A particle moves according to the law,

x = rcos\frac{{\pi t}}{{2.}}

The distance covered by it the time interval between t= 0 to t=3s is

View answer and explanation

Correct option: C

3r

Option C (3r) is correct because the particle travels a distance of 'r' from t=0 to t=1, then returns 'r' from t=1 to t=2, and then travels another 'r' from t=2 to t=3, resulting in a total distance of 3r.

Solution approach

Correct Answer + Core Reason: $3r$ . The distance covered is the sum of the absolute values of the displacements between the given times.

Step-by-Step Solution:

Step 1: Determine the position at $t=0$ : $x(0) = r \cos(\frac{\pi (0)}{2}) = r \cos(0) = r$ .

Step 2: Determine the position at $t=1$ : $x(1) = r \cos(\frac{\pi (1)}{2}) = r \cos(\frac{\pi}{2}) = 0$ .

Step 3: Determine the position at $t=2$ : $x(2) = r \cos(\frac{\pi (2)}{2}) = r \cos(\pi) = -r$ .

Step 4: Determine the position at $t=3$ : $x(3) = r \cos(\frac{\pi (3)}{2}) = r \cos(\frac{{3\pi}}{2}) = 0$ .

Step 5: Calculate the distance from $t=0$ to $t=1$ : $|x(1) - x(0)| = |0 - r| = r$ .

Step 6: Calculate the distance from $t=1$ to $t=2$ : $|x(2) - x(1)| = |-r - 0| = r$ .

Step 7: Calculate the distance from $t=2$ to $t=3$ : $|x(3) - x(2)| = |0 - (-r)| = r$ .

Step 8: Calculate the total distance: $r + r + r = 3r$ .

Why other options are wrong:
- **Option A ( $r$ ): This is incorrect because it only considers the distance from t=0 to t=1, but not the subsequent movements.
- Option B ( $2r$ ): This is incorrect because it doesn't account for the movement from t=2 to t=3.
- Option D ( $4r$ ): This is incorrect as the particle doesn't cover a distance of 4r in the given time interval.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Distance vs. Displacement

Memory Aids:
- Distance is the total path length, while displacement is the change in position.

Quick Tips:**
- Pay attention to the difference between distance and displacement. In SHM, the distance can be greater than the amplitude.

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Understanding +4 | -1

(A): Since and cosine functions are periodic functions
(R): Sinusoidal functions repeat its values after a definite interval of time

View answer and explanation

Correct option: A

If Both A

\&

R are true and R is correct explanation of A

Option A (If Both A & R are true and R is correct explanation of A) is correct because both the assertion and the reason are true, and the reason correctly explains why sine and cosine functions are periodic.

Solution approach

Correct Answer + Core Reason: If Both A $\& R are true and R is correct explanation of A. Both the assertion and reason are true, and the reason correctly explains the assertion.\n\n**Step-by-Step Solution:**\n\n**Step 1:** Analyze the Assertion (A): Sine and cosine functions are periodic functions. This statement is true because both sine and cosine functions repeat their values after a fixed interval.\n\n**Step 2:** Analyze the Reason (R): Sinusoidal functions repeat their values after a definite interval of time. This statement is also true and explains why sine and cosine functions are periodic.\n\n**Step 3:** Determine if R is the correct explanation for A: The reason (R) directly explains why the assertion (A) is true. The periodicity of sine and cosine functions is due to their repeating values after a definite interval of time.\n\n**Why other options are wrong:**\n- **Option B:** If Both A$ \& R are true but R is not correct explanation of A: This is incorrect because R is the correct explanation for A.
- Option C: If A is true but R is false: This is incorrect because both A and R are true.
- Option D: If both A $\& R are false: This is incorrect because both A and R are true.\n\n**Key Concepts:**\n- Periodic functions\n- Sine and cosine functions\n- Periodicity\n\n**Memory Aids:**\n- Remember that periodic functions repeat their values after a fixed interval.\n\n**Quick Tips:**\n- Understand the definitions of assertion and reason questions.\n- Check if both statements are true and if the reason correctly explains the assertion.$

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Application +4 | -1

Velocity of a simple harmonic oscillation is given by

V = \sqrt {144 - 16{x^2}} {\mkern 1mu} {\mkern 1mu} m{s^{ - 1}}

where x is displacement in metre. Then maximum velocity of that oscillator is

View answer and explanation

Correct option: B

12\,\,m{s^{ - 1}}

Option B (

12\,m{s^{ - 1}}

) is correct because the maximum velocity of the oscillator is found by comparing the given equation with the standard form of velocity in SHM.

Solution approach

Correct Answer + Core Reason: $12\,m{s^{ - 1}}$ . The maximum velocity is found by comparing the given equation with the standard form.

Step-by-Step Solution:

Step 1: Identify the given equation for velocity: $V = \sqrt{144 - 16x^2} \, m/s$ .

Step 2: Rewrite the equation in a more recognizable form. Factor out 16 from under the square root: $V = \sqrt{16(9 - x^2)} = 4\sqrt{9 - x^2} = 4\sqrt{3^2 - x^2}$ .

Step 3: Recall the standard form of velocity in SHM: $v = \omega \sqrt{A^2 - x^2}$ , where $\omega$ is the angular frequency and $A$ is the amplitude.

Step 4: Compare the given equation with the standard form. We have $\omega = 4$ and $A = 3$ . Therefore, $V = 4\sqrt{3^2 - x^2}$ .

Step 5: The maximum velocity occurs when $x = 0$ . So, $V_{max} = \omega A = 4 \times 3 = 12 \, m/s$ .

Why other options are wrong:
- Option A: $6\,m{s^{ - 1}}$ is incorrect as it does not match the correct calculation.
- Option C: $16\,m{s^{ - 1}}$ is incorrect as it does not match the correct calculation.
- Option D: $8\,m{s^{ - 1}}$ is incorrect as it does not match the correct calculation.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Velocity in SHM
- Amplitude
- Angular frequency

Memory Aids:
- Remember the formula for velocity in SHM: $v = \omega \sqrt{A^2 - x^2}$
- Maximum velocity occurs at the equilibrium position (x=0) and is given by $v_{max} = \omega A$

Quick Tips:
- Recognize the standard form of equations in SHM to easily identify parameters like amplitude and angular frequency.
- The maximum velocity occurs when the displacement is zero.

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Remembering +4 | -1

Which one of the following is the correct relation between frequency

\mathrm{f}

and angular frequency

\omega

?

View answer and explanation

Correct option: B

\omega=2 \pi \mathrm{f}

Option B (

\omega=2 \pi \mathrm{f}

) is correct because it is the fundamental relationship between angular frequency and frequency.

Solution approach

Correct Answer + Core Reason: $\omega=2 \pi \mathrm{f}$ . This is the correct relationship between angular frequency and frequency.

Step-by-Step Solution:

Step 1: Recall the definition of frequency (f): It is the number of oscillations per unit time (usually seconds).

Step 2: Recall the definition of angular frequency ( $\omega$ ): It is the rate of change of the angle of an object moving on a circular path, or the rate of change of the phase of a sinusoidal waveform.

Step 3: Relate the two quantities. One complete oscillation corresponds to an angle of $2\pi$ radians. Therefore, the angular frequency is $2\pi$ times the frequency: $\omega = 2\pi f$ .

Why other options are wrong:
- **Option A ( $\mathrm{f}=\pi \omega$ ):** This is incorrect because it has the wrong relationship between f and $\omega$ .
- **Option C ( $\mathrm{f}=2 \omega / \pi$ ):** This is incorrect because it has the wrong relationship between f and $\omega$ .
- **Option D ( $\mathrm{f}=2 \pi \omega$ ):** This is incorrect because it has the wrong relationship between f and $\omega$ .

Key Concepts:
- Frequency
- Angular Frequency
- Relationship between Frequency and Angular Frequency

Memory Aids:
- Remember the formula $\omega = 2\pi f$ .

Quick Tips:
- Always use the correct relationship between frequency and angular frequency when solving problems involving oscillations.

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Analyzing +4 | -1

The bob of a pendulum is in the lowest position and has a horizontal velocity of

\sqrt {gL}

where

L

is length of the pendulum and

g

is acceleration due to gravity is given to it. Then the bob

View answer and explanation

Correct option: C

Oscillates about the lowest position as mean position

Option C (Oscillates about the lowest position as mean position) is correct because the initial velocity is not high enough for the bob to complete a full circle.

Solution approach

Correct Answer + Core Reason: Oscillates about the lowest position as mean position. The bob oscillates because the initial velocity is less than the velocity required to complete a vertical circle.

Step-by-Step Solution:

Step 1: The minimum velocity required to complete a vertical circle at the lowest point is $v = \sqrt{5gL}$ .

Step 2: The given velocity is $v = \sqrt{gL}$ .

Step 3: Since $\sqrt{gL} < \sqrt{5gL}$ , the bob will not complete a vertical circle.

Step 4: The bob will rise to a certain height and then oscillate about the lowest position as the mean position.

Why other options are wrong:
- Option A: Just completes a vertical circle. This is incorrect because the initial velocity is less than the required velocity.
- Option B: Just completes semi-circle. This is incorrect because the initial velocity is not sufficient to reach the top of the circle (semi-circle).
- Option D: Oscillates as conceal pendulum. This is incorrect because the pendulum oscillates in a vertical plane, not as a conical pendulum.

Key Concepts:
- Vertical Circular Motion
- Simple Pendulum
- Conservation of Energy

Memory Aids:
- Remember the minimum velocity required to complete a vertical circle at the lowest point: $v = \sqrt{5gL}$ .

Quick Tips:
- Compare the given velocity with the minimum velocity required to complete a vertical circle.
- If the given velocity is less than the minimum velocity, the bob will oscillate.

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Remembering +4 | -1

The differential equation of a particle executing simple harmonic motion along

y

-axis is

View answer and explanation

Correct option: A

\frac{d^2 y}{d t^2}+\omega^2 y=0

Option A (

\frac{{d^2 y}}{{d t^2}}+\omega^2 y=0

) is correct because it represents the standard differential equation for SHM.

Solution approach

Correct Answer + Core Reason: $\frac{{d^2 y}}{{d t^2}}+\omega^2 y=0$ . This is the standard differential equation for SHM.

Step-by-Step Solution:

Step 1: Recall the general equation for SHM: $y(t) = A \cos(\omega t + \phi)$ , where A is the amplitude, $\omega$ is the angular frequency, and $\phi$ is the phase constant.

Step 2: Find the first derivative of y with respect to time: $\frac{{dy}}{{dt}} = -A\omega \sin(\omega t + \phi)$ .

Step 3: Find the second derivative of y with respect to time: $\frac{{d^2y}}{{dt^2}} = -A\omega^2 \cos(\omega t + \phi)$ .

Step 4: Substitute the original equation $y(t) = A \cos(\omega t + \phi)$ into the second derivative equation: $\frac{{d^2y}}{{dt^2}} = -\omega^2 y$ .

Step 5: Rearrange the equation to get the standard form of the differential equation for SHM: $\frac{{d^2y}}{{dt^2}} + \omega^2 y = 0$ .

Why other options are wrong:
- **Option B ( $\frac{{d^2 y}}{{d t^2}}+\omega^2 y^2=0$ ):** This is incorrect because it has a $y^2$ term, which is not part of the standard SHM differential equation.
- **Option C ( $\frac{{d^2 y}}{{d t^2}}-\omega^2 y=0$ ):** This is incorrect because it has a negative sign in front of the $\omega^2 y$ term. This would represent exponential growth or decay, not oscillation.
- **Option D ( $\frac{{d y}}{{d t}}+\omega y=0$ ): This is incorrect because it's a first-order differential equation, not a second-order one, and it doesn't represent SHM.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Differential Equation of SHM

Memory Aids:**
- Remember the standard form of the differential equation for SHM: $\frac{{d^2y}}{{dt^2}} + \omega^2 y = 0$ .

Quick Tips:
- The differential equation for SHM is a second-order linear homogeneous differential equation with constant coefficients.

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Analyzing +4 | -1

The displacement

y

of a particle is given by

y=4 \cos ^2\left(\frac{t}{2}\right) \sin (1000 t)

. This expression may be considered to be a result of the superposition of how many simple harmonic motion

View answer and explanation

Correct option: A

3

Option A (3) is correct because the given expression can be simplified using trigonometric identities to a superposition of three simple harmonic motions.

Solution approach

Correct Answer + Core Reason: 3. The given expression can be simplified to a sum of three SHM terms using trigonometric identities.

Step-by-Step Solution:

Step 1: Use the trigonometric identity $\cos^2(x) = \frac{1 + \cos(2x)}{2}$ to rewrite the given expression:
$y = 4 \cos^2(\frac{t}{2}) \sin(1000t) = 4 \left( \frac{1 + \cos(t)}{2} \right) \sin(1000t) = 2(1 + \cos(t)) \sin(1000t)$ .

Step 2: Expand the expression: $y = 2\sin(1000t) + 2\cos(t)\sin(1000t)$ .

Step 3: Use the trigonometric identity $2\cos(A)\sin(B) = \sin(A+B) - \sin(A-B)$ to rewrite the second term:
$2\cos(t)\sin(1000t) = \sin(1000t + t) + \sin(1000t - t) = \sin(1001t) - \sin(-999t) = \sin(1001t) + \sin(999t)$ .

Step 4: Substitute this back into the expression for $y$ : $y = 2\sin(1000t) + \sin(1001t) + \sin(999t)$ .

Step 5: Count the number of simple harmonic motions: There are three terms, each representing a simple harmonic motion with different frequencies (1000, 1001, and 999).

Why other options are wrong:
- Option B (0): This is incorrect because the expression can be simplified into three SHM terms.
- Option C (0): This is incorrect because the expression can be simplified into three SHM terms.
- Option D (0): This is incorrect because the expression can be simplified into three SHM terms.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Trigonometric Identities
- Superposition of SHM

Memory Aids:
- Remember the trigonometric identities for $\cos^2(x)$ and $2\cos(A)\sin(B)$ .

Quick Tips:
- Use trigonometric identities to simplify complex expressions into a sum of simple harmonic motions.

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Application +4 | -1

An object on a spring vibrates in a simple harmonic motion at a frequency of 4 hertz and an amplitude of 8 cm. If the mass of the object is 200 g, the spring constant is

View answer and explanation

Correct option: C

126 N/m

Option C (126 N/m) is correct because it is calculated using the frequency and mass to find the spring constant.

Solution approach

Correct Answer + Core Reason: 126 N/m. The spring constant is calculated from the frequency and mass.

Step-by-Step Solution:

Step 1: Given: frequency $f = 4 \mathrm{~Hz}$ , amplitude $A = 8 \mathrm{~cm} = 0.08 \mathrm{~m}$ , mass $m = 200 \mathrm{~g} = 0.2 \mathrm{~kg}$ .

Step 2: The angular frequency is $\omega = 2\pi f = 2\pi (4) = 8\pi \mathrm{~rad/s}$ .

Step 3: The spring constant $k$ is related to the angular frequency and mass by $\omega = \sqrt{\frac{k}{m}}$ .

Step 4: Solve for $k$ : $k = m\omega^2 = 0.2 \times (8\pi)^2 = 0.2 \times 64 \pi^2 \approx 0.2 \times 64 \times 9.87 \approx 126.336 \mathrm{~N/m} \approx 126 \mathrm{~N/m}$

Why other options are wrong:
- Option A: Incorrect. This value doesn't match the correct calculation using the given frequency and mass.
- Option B: Incorrect. This value doesn't match the correct calculation using the given frequency and mass.
- Option D: Incorrect. This value doesn't match the correct calculation using the given frequency and mass.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Angular frequency: $\omega = 2\pi f$
- Relationship between spring constant, mass, and angular frequency: $\omega = \sqrt{\frac{k}{m}}$

Memory Aids:
- Remember the formula $\omega = \sqrt{\frac{k}{m}}$ .

Quick Tips:
- Make sure to convert all the given values to SI units before plugging them into the formulas.

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Understanding +4 | -1

The average acceleration of a particle performing SHM over one complete oscillation is

View answer and explanation

Correct option: C

Zero

Option C (Zero) is correct because the velocity changes direction over the oscillation, resulting in zero average acceleration.

Solution approach

Correct Answer + Core Reason: Zero. Over a complete oscillation, the velocity changes direction, resulting in zero average acceleration.

Step-by-Step Solution:

Step 1: Define average acceleration: Average acceleration is the change in velocity divided by the change in time: $a_{avg} = \frac{\Delta v}{\Delta t}$ .

Step 2: In SHM, after one complete oscillation, the particle returns to its starting point with the same speed but in the opposite direction (if starting at an extreme) or the same speed and direction (if starting at the mean position).

Step 3: If the particle starts at an extreme position, its velocity at the beginning is 0 and its velocity after one oscillation is also 0. If the particle starts at the mean position, its velocity is +v and after one oscillation, its velocity is also +v. In either case, the *change* in velocity over one full oscillation, considering direction, is zero.

Step 4: Therefore, the average acceleration over one complete oscillation is zero.

Why other options are wrong:
- Option A: $\frac{{\omega ^2}A}{2}$ . Incorrect, this is not the average acceleration.
- Option B: $\frac{{\omega ^2}A}{{\sqrt 2 }}$ . Incorrect, this is not the average acceleration.
- Option D: $A{\omega ^2}$ . Incorrect, this is the maximum acceleration, not the average.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Average acceleration
- Oscillation

Memory Aids:
- Average acceleration over a cycle is zero.

Quick Tips:
- Consider the direction of velocity when calculating average acceleration.

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Application +4 | -1

The time period of a particle performing linear SHM is 18s. What is the time taken by it to make a displacement equal to half its amplitude from its equilibrium position ?

View answer and explanation

Correct option: C

\frac{3}{2}s

Option C (

\frac{3}{2}s

) is correct because it is calculated using the SHM equation and the given time period.

Solution approach

Correct Answer + Core Reason: $\frac{3}{2}s$ . The time taken to reach half the amplitude is calculated using the SHM equation and the given time period.

Step-by-Step Solution:

Step 1: Define the SHM equation: $x = A \sin(\omega t)$ , where $x$ is the displacement, $A$ is the amplitude, $\omega$ is the angular frequency, and $t$ is the time.

Step 2: We are given that $x = \frac{A}{2}$ . Substitute this into the equation: $\frac{A}{2} = A \sin(\omega t)$ .

Step 3: Simplify the equation: $\frac{1}{2} = \sin(\omega t)$ .

Step 4: Find the angle whose sine is $\frac{1}{2}$ : $\omega t = \frac{\pi}{6}$ .

Step 5: We know that $\omega = \frac{2\pi}{T}$ , where $T$ is the time period. Given $T = 18s$ , we have $\omega = \frac{2\pi}{18} = \frac{\pi}{9}$ .

Step 6: Substitute the value of $\omega$ into the equation $\omega t = \frac{\pi}{6}$ : $\frac{\pi}{9} t = \frac{\pi}{6}$ .

Step 7: Solve for $t$ : $t = \frac{\pi}{6} \times \frac{9}{\pi} = \frac{9}{6} = \frac{3}{2}s$ .

Final Answer: $t = \frac{3}{2}s$

Why other options are wrong:
- Option A (3s): This is incorrect because it doesn't correctly account for the sinusoidal relationship between displacement and time in SHM.
- Option B (2s): This is incorrect because it doesn't correctly account for the sinusoidal relationship between displacement and time in SHM.
- Option D (1s): This is incorrect because it doesn't correctly account for the sinusoidal relationship between displacement and time in SHM.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Angular frequency
- Time period
- Sinusoidal relationship in SHM

Memory Aids:
- Remember the SHM equation: $x = A \sin(\omega t)$
- $\omega = \frac{2\pi}{T}$

Quick Tips:
- Always remember the standard SHM equation.
- Convert time period to angular frequency correctly.
- Remember the values of sine for standard angles.

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Remembering +4 | -1

Maximum acceleration in SHM

View answer and explanation

Correct option: A

\omega^2 A

Option A (

\omega^2 A

) is correct because the maximum acceleration in SHM is given by the formula

a_{max} = \omega^2 A

.

Solution approach

Correct Answer + Core Reason: $\omega^2 A$ . The maximum acceleration in SHM is given by $a_{max} = \omega^2 A$ .

Step-by-Step Solution:

Step 1: Recall the equation for displacement in SHM: $x(t) = A \cos(\omega t + \phi)$ , where A is the amplitude, $\omega$ is the angular frequency, and $\phi$ is the phase constant.

Step 2: Find the velocity by taking the first derivative of the displacement with respect to time: $v(t) = \frac{{dx}}{{dt}} = -A\omega \sin(\omega t + \phi)$ .

Step 3: Find the acceleration by taking the second derivative of the displacement with respect to time: $a(t) = \frac{{dv}}{{dt}} = \frac{{d^2x}}{{dt^2}} = -A\omega^2 \cos(\omega t + \phi)$ .

Step 4: Determine the maximum acceleration. The maximum value of the acceleration occurs when $|\cos(\omega t + \phi)| = 1$ . Therefore, $a_{max} = A\omega^2 = \omega^2 A$ .

Why other options are wrong:
- **Option B ( $\frac{{\omega^2 A}}{2}$ ): This is incorrect because it's half the maximum acceleration.
- Option C ( $\omega^2 A^2$ ): This is incorrect because it has the wrong units and is not the correct formula.
- Option D (0): This is incorrect because the acceleration is not always zero in SHM. It is zero at the equilibrium position, but maximum at the extreme positions.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Displacement, Velocity, and Acceleration in SHM

Memory Aids:
- Remember the relationship between displacement, velocity, and acceleration in SHM.

Quick Tips:**
- The maximum acceleration occurs at the extreme positions, while the maximum velocity occurs at the equilibrium position.

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Application +4 | -1

The bob of a simple pendulum of mass m is displaced from its equilibrium position through height ‘h’ and then released. It vibrates with a time period of 2 seconds. The tension in the bob when the bob passes through mean position

View answer and explanation

Correct option: A

m\left( {g + 2{\pi ^2}h} \right)

Option A (

m(g + 2{\pi ^2}h)

) is correct because the tension at the mean position is the sum of the weight and the centripetal force, which depends on the height 'h' and the time period.

Solution approach

Correct Answer + Core Reason: $m\left( {g + 2{\pi ^2}h} \right)$ . The tension in the bob at the mean position is the sum of the weight and the centripetal force.

Step-by-Step Solution:

Step 1: The bob is released from a height $h$ . At the mean position, all the potential energy is converted to kinetic energy.

Step 2: Using conservation of energy, $mgh = \frac{1}{2}mv^2$ , where $v$ is the velocity at the mean position.

Step 3: Solving for $v^2$ , we get $v^2 = 2gh$ .

Step 4: At the mean position, the net force towards the center is the tension $T$ minus the weight $mg$ . This net force provides the centripetal force: $T - mg = \frac{mv^2}{L}$ , where $L$ is the length of the pendulum.

Step 5: Substituting $v^2 = 2gh$ , we get $T - mg = \frac{m(2gh)}{L}$ .

Step 6: Therefore, $T = mg + \frac{2mgh}{L}$ .

Step 7: The time period of a simple pendulum is $T = 2\pi \sqrt{\frac{L}{g}}$ . We are given that $T = 2$ seconds, so $2 = 2\pi \sqrt{\frac{L}{g}}$ .

Step 8: Squaring both sides, we get $4 = 4\pi^2 \frac{L}{g}$ , so $L = \frac{g}{\pi^2}$ .

Step 9: Substituting $L = \frac{g}{\pi^2}$ into the expression for tension, we get $T = mg + \frac{2mgh}{\frac{g}{\pi^2}} = mg + 2mh\pi^2 = m(g + 2\pi^2 h)$ .

Why other options are wrong:
- **Option B ( $m\left( {g + {\pi ^2}h} \right)$ ): This is incorrect because the centripetal force is not calculated correctly.
- Option C ( $m\left( {g + \frac{{\pi ^2}}{2}h} \right)$ ): This is incorrect because the centripetal force is not calculated correctly.
- Option D ( $m\left( {g + \frac{{\pi ^2}}{3}h} \right)$ ): This is incorrect because the centripetal force is not calculated correctly.

Key Concepts:
- Simple pendulum
- Conservation of energy
- Centripetal force
- Tension

Memory Aids:**
- $T = 2\pi \sqrt{\frac{L}{g}}$
- $KE = \frac{1}{2}mv^2$
- $PE = mgh$

Quick Tips:
- Remember to use conservation of energy to find the velocity at the mean position.
- The tension at the mean position is the sum of the weight and the centripetal force.

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Understanding +4 | -1

If a body is executing simple harmonic motion then

View answer and explanation

Correct option: C

At equilibrium position, the total energy is in the form of kinetic energy

Option C (At equilibrium position, the total energy is in the form of kinetic energy) is correct because at the equilibrium position, the potential energy is zero, and all the energy is kinetic.

Solution approach

Correct Answer + Core Reason: At equilibrium position, the total energy is in the form of kinetic energy. At equilibrium, potential energy is zero, so total energy is kinetic.

Step-by-Step Solution:

Step 1: Total energy in SHM: $E = \frac{{1}}{{2}}kA^2$ , where k is the spring constant and A is the amplitude.

Step 2: Potential energy in SHM: $U = \frac{{1}}{{2}}kx^2$ , where x is the displacement from the mean position.

Step 3: Kinetic energy in SHM: $K = \frac{{1}}{{2}}mv^2$ , where v is the velocity.

Step 4: At the equilibrium position, x = 0, so the potential energy U = 0.

Step 5: Therefore, at the equilibrium position, the total energy is equal to the kinetic energy: $E = K$ .

Why other options are wrong:
- Option A: Incorrect. At extreme positions, the total energy is not zero; it's in the form of potential energy.
- Option B: Incorrect. At the equilibrium position, the total energy is in the form of kinetic energy, not potential energy.
- Option D: Incorrect. The total energy is finite and equal to $\frac{{1}}{{2}}kA^2$ .

Key Concepts:
- Simple Harmonic Motion (SHM)
- Total Energy
- Potential Energy
- Kinetic Energy
- Equilibrium Position
- Extreme Position

Memory Aids:
- Remember that at the equilibrium position, potential energy is zero, and at the extreme positions, kinetic energy is zero.

Quick Tips:
- Visualize the energy exchange between potential and kinetic energy in SHM.

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Understanding +4 | -1

Consider the following statements : The total energy of a particle executing simple harmonic motion depends on its I. Amplitude II. Period III. Displacement Of these statements

View answer and explanation

Correct option: A

I and II are correct

Option A (I and II are correct) is correct because the total energy in SHM depends on amplitude and period, but not instantaneous displacement.

Solution approach

Correct Answer + Core Reason: I and II are correct. Total energy depends on amplitude and period (or frequency) but not displacement.

Step-by-Step Solution:

Step 1: Recall the formula for the total energy of a particle executing SHM: $E = \frac{1}{2} m \omega^2 A^2$ , where $m$ is the mass, $\omega$ is the angular frequency, and $A$ is the amplitude.

Step 2: Express angular frequency in terms of the period: $\omega = \frac{2\pi}{T}$ .

Step 3: Substitute this into the energy formula: $E = \frac{1}{2} m (\frac{2\pi}{T})^2 A^2 = \frac{2\pi^2 m A^2}{T^2}$ .

Step 4: From the formula, it's clear that the total energy depends on the amplitude ( $A$ ) and the period ( $T$ ). It does not depend on the instantaneous displacement.

Why other options are wrong:
- Option B: II and III are correct. Incorrect, energy does not depend on displacement.
- Option C: I and III are correct. Incorrect, energy does not depend on displacement.
- Option D: I , II and III are correct. Incorrect, energy does not depend on displacement.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Total energy in SHM
- Amplitude
- Period
- Displacement

Memory Aids:
- Energy is proportional to amplitude squared and inversely proportional to period squared.

Quick Tips:
- Remember the energy formula for SHM.

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Analyzing +4 | -1

The bob of a simple pendulum is a spherical hollow ball filled with water. A plugged hole near the bottom of the oscillating bob gets suddenly unplugged. During observation, till water is coming out, the time period of oscillation would

View answer and explanation

Correct option: C

First increase and then decrease to the original value

Option C (First increase and then decrease to the original value) is correct because as water flows out, the center of mass initially lowers, increasing the effective length, and then rises, decreasing the effective length, returning to the original value.

Solution approach

Correct Answer + Core Reason: First increase and then decrease to the original value. As water flows out, the center of mass initially lowers, then rises, affecting the period of oscillation.

Step-by-Step Solution:

Step 1: Initially, the bob is a hollow ball filled with water. The center of mass is at the center of the sphere.

Step 2: As water starts flowing out, the center of mass of the bob + water system initially goes down (lower than the center of the sphere). This is because the water is leaving from the bottom.

Step 3: As more water flows out, the center of mass eventually starts rising back towards the center of the sphere, ultimately returning to the center of the sphere when all the water is gone.

Step 4: The time period of oscillation of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$ , where $L$ is the effective length (distance between the point of suspension and the center of mass).

Step 5: Since the center of mass initially goes down, $L$ increases, and the time period $T$ increases. Then, as the center of mass goes up, $L$ decreases, and the time period $T$ decreases. Finally, when all the water is gone, the center of mass is back at the center of the sphere, and the time period returns to its original value.

Why other options are wrong:
- Option A: Remain unchanged is incorrect because the center of mass changes, affecting the time period.
- Option B: First decrease and then increase to the original value is incorrect because the center of mass initially goes down, causing the period to increase first.
- Option D: Increase towards a saturation value is incorrect because the period eventually returns to its original value.

Key Concepts:
- Time period of a simple pendulum: $T = 2\pi \sqrt{\frac{L}{g}}$
- Center of mass

Memory Aids:
- Visualize the movement of the center of mass as water flows out.

Quick Tips:
- Consider how the center of mass changes as the water level decreases.

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Application +4 | -1

A point performs simple harmonic oscillation of period T and the equation of motion is given by

x = a\,\,\,\sin \,\,\left( {w + \pi /6} \right)

After the elapse of what fraction of the time period the velocity of the point will be equal to half its maximum velocity

View answer and explanation

Correct option: B

\frac{T}{{12}}

Option B (

\frac{T}{{12}}

) is correct because it involves finding the time when velocity is half of maximum velocity in SHM.

Solution approach

Correct Answer + Core Reason: $\frac{T}{{12}}$ . The time is calculated by finding when the velocity is half its maximum value in SHM.

Step-by-Step Solution:

Step 1: Given: $x = a \sin(\omega t + \frac{\pi}{6})$ .

Step 2: Velocity $v = \frac{dx}{dt} = a\omega \cos(\omega t + \frac{\pi}{6})$ .

Step 3: Maximum velocity $v_{max} = a\omega$ .

Step 4: We want to find $t$ such that $v = \frac{1}{2}v_{max}$ , so $\frac{1}{2}a\omega = a\omega \cos(\omega t + \frac{\pi}{6})$ .

Step 5: This simplifies to $\cos(\omega t + \frac{\pi}{6}) = \frac{1}{2}$ .

Step 6: Therefore, $\omega t + \frac{\pi}{6} = \frac{\pi}{3}$ .

Step 7: Solving for $t$ , we get $\omega t = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$ .

Step 8: Since $\omega = \frac{2\pi}{T}$ , we have $\frac{2\pi}{T}t = \frac{\pi}{6}$ .

Step 9: Solving for $t$ , we get $t = \frac{T}{12}$ .

Why other options are wrong:
- Option A: $\frac{T}{3}$ This is incorrect. It might be obtained by incorrectly solving the trigonometric equation.
- Option C: $\frac{T}{8}$ This is incorrect. It might be obtained by incorrectly solving the trigonometric equation.
- Option D: $\frac{T}{6}$ This is incorrect. It might be obtained by incorrectly solving the trigonometric equation.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Velocity in SHM
- Angular Frequency

Memory Aids:
- Remember the relationship between angular frequency and time period: $\omega = \frac{2\pi}{T}$ .

Quick Tips:
- Differentiate the displacement equation to find the velocity equation.
- Solve the trigonometric equation carefully.

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Application +4 | -1

Two simple pendulums first of bob mass

{M_1}

and length

{L_1}

second of bob mass

{M_2}

and length

{L_2}

.

{M_1}

=

{M_2}

and

{L_1}

=

{2L_2}

. If the vibrational energy of both is same. Then which is correct

View answer and explanation

Correct option: B

Amplitude of B smaller than A

Option B, Amplitude of B smaller than A, is correct because for the same vibrational energy, the pendulum with a shorter length (and thus higher frequency) will have a smaller amplitude.

Solution approach

Correct Answer + Core Reason: Amplitude of B smaller than A. For equal vibrational energy, shorter pendulum (higher frequency) has smaller amplitude.

Step-by-Step Solution:

Step 1: Write the expression for the potential energy of a simple pendulum: $U = \frac{1}{2}m{\omega^2}{A^2}$ , where $m$ is mass, $\omega$ is angular frequency, and $A$ is amplitude.

Step 2: Since the vibrational energy is the same for both pendulums, $U_1 = U_2$ .

Step 3: Write the expression for angular frequency: $\omega = \sqrt{\frac{g}{L}}$ , where $g$ is acceleration due to gravity and $L$ is the length of the pendulum.

Step 4: Substitute the angular frequency into the potential energy expression: $U = \frac{1}{2}m\frac{g}{L}{A^2}$ .

Step 5: Since $U_1 = U_2$ , we have $\frac{1}{2}{M_1}\frac{g}{{{L_1}}}{A_1}^2 = \frac{1}{2}{M_2}\frac{g}{{{L_2}}}{A_2}^2$ .

Step 6: Given ${M_1} = {M_2}$ and ${L_1} = 2{L_2}$ , substitute these values into the equation: $\frac{1}{{{L_1}}}{A_1}^2 = \frac{1}{{{L_2}}}{A_2}^2 \Rightarrow \frac{1}{{2{L_2}}}{A_1}^2 = \frac{1}{{{L_2}}}{A_2}^2$ .

Step 7: Simplify the equation: ${A_1}^2 = 2{A_2}^2 \Rightarrow {A_1} = \sqrt{2} {A_2}$ . Therefore, ${A_2} = \frac{{{A_1}}}{{\sqrt{2} }}$ .

Step 8: Conclude that the amplitude of pendulum B ( $A_2$ ) is smaller than the amplitude of pendulum A ( $A_1$ ).

Why other options are wrong:
- Option A: Amplitude of B greater than A. This is incorrect because the shorter pendulum (B) has a higher frequency, and for the same energy, it must have a smaller amplitude.
- Option C: Amplitude will be same. This is incorrect because the lengths of the pendulums are different, leading to different frequencies and therefore different amplitudes for the same energy.
- Option D: None of these. This is incorrect because one of the options is correct.

Key Concepts:
- Simple Pendulum
- Potential Energy of a Simple Pendulum
- Angular Frequency

Memory Aids:
- $\omega = \sqrt{\frac{g}{L}}$ (Angular frequency of a simple pendulum)
- $U = \frac{1}{2}m{\omega^2}{A^2}$ (Potential energy in SHM)

Quick Tips:
- Remember that for the same energy, a higher frequency corresponds to a smaller amplitude in SHM.
- The length of the pendulum affects its frequency, which in turn affects the amplitude for a given energy.

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Analyzing +4 | -1

A particle executes SHM, the graph of velocity as a function of displacement is

View answer and explanation

Correct option: C

an ellipse

Option C (an ellipse) is correct because the relationship between velocity and displacement in SHM is described by an ellipse equation.

Solution approach

Correct Answer + Core Reason: an ellipse. The graph of velocity as a function of displacement in SHM is an ellipse.

Step-by-Step Solution:

Step 1: Write the equation for displacement in SHM:\nx = A \cos(ω t)

Step 2: Write the equation for velocity in SHM:\nv = -Aω \sin(ω t)

Step 3: Express \sin(ω t) and \cos(ω t) in terms of x and v:
\cos(ω t) = $\frac{x}{A}$
\sin(ω t) = - $\frac{v}{Aω}$

Step 4: Use the trigonometric identity $\sin^2(\omega t) + \cos^2(\omega t) = 1$ :
$\left(\frac{x}{A}\right)^2 + \left(-\frac{v}{A\omega}\right)^2 = 1$
$\frac{x^2}{A^2} + \frac{v^2}{A^2\omega^2} = 1$

Step 5: Recognize the equation as the equation of an ellipse:\nThis equation represents an ellipse in the x-v plane, where the semi-major and semi-minor axes are A and A $\omega$ , respectively.

Why other options are wrong:
- Option A: A circle would require the coefficients of $x^2$ and $v^2$ to be equal, which is not the case here.
- Option B: A parabola represents a quadratic relationship, which is not present between velocity and displacement in SHM.
- Option D: A helix is a three-dimensional curve, while the relationship between velocity and displacement is two-dimensional.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Displacement in SHM
- Velocity in SHM
- Equation of an ellipse

Memory Aids:
- Remember the standard equations for displacement and velocity in SHM.
- Recall the trigonometric identity $\sin^2(\theta) + \cos^2(\theta) = 1$ .

Quick Tips:
- When asked about the graph of velocity vs displacement in SHM, remember it's always an ellipse.
- The shape of the ellipse depends on the amplitude and angular frequency of the SHM.

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Analyzing +4 | -1

The displacement of a particle along the

{x}

-axis is given by The motion of the particle corresponds to

View answer and explanation

Correct option: C

Non simple harmonic motion

Option C (Non simple harmonic motion) is correct because the given displacement equation is not in the form of a simple harmonic function (sine or cosine).

Solution approach

Correct Answer + Core Reason: Non simple harmonic motion. The given displacement equation is not in the form of a simple harmonic function.

Step-by-Step Solution:

Step 1: The displacement of a particle is given by $x = A\cos(\omega t) + B\cos(3\omega t)$ .

Step 2: For simple harmonic motion, the displacement should be of the form $x = C\cos(\omega t + \phi)$ or $x = C\sin(\omega t + \phi)$ , where $C$ is the amplitude and $\phi$ is the phase constant.

Step 3: The given equation is a superposition of two cosine functions with different frequencies ( $\omega$ and $3\omega$ ). Therefore, it is periodic but not simple harmonic.

Why other options are wrong:
- Option A: Simple harmonic motion of frequency $\omega / \pi$ is incorrect because the motion is not simple harmonic.
- Option B: Simple harmonic motion of frequency 3 omega / $2 \pi$ is incorrect because the motion is not simple harmonic.
- Option D: Simple harmonic motion of frequency $\omega / 2 \pi$ is incorrect because the motion is not simple harmonic.

Key Concepts:
- Simple harmonic motion (SHM)
- Superposition of waves

Memory Aids:
- Remember that SHM can be represented by a simple sine or cosine function.

Quick Tips:
- Check if the given equation can be reduced to a simple sine or cosine function to determine if it represents SHM.

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Understanding +4 | -1

A particle executes simple harmonic motion between

x

= - A\,\,\mathrm{and}\,\,x = + A

The time taken for it to go from

0\mathrm{ }to\;A/2\,\,is\,\,{T_1}

and to go from

A/2

to

A\,\,\mathrm{is}\,\,{T_2}

Then

View answer and explanation

Correct option: A

{T_1} < {T_2}

Option A (

{T_1} < {T_2}

) is correct because the speed of the particle is higher near the mean position (x=0) and lower near the extreme position (x=A).

Solution approach

Correct Answer + Core Reason: ${T_1} < {T_2}$ . The particle's speed is higher near the mean position.

Step-by-Step Solution:

Step 1: Recall that in SHM, the velocity is maximum at the mean position (x = 0) and minimum at the extreme positions (x = ±A).

Step 2: The particle travels from 0 to A/2 in time $T_1$ and from A/2 to A in time $T_2$ .

Step 3: Since the velocity is higher between 0 and A/2 than between A/2 and A, it will take less time to cover the distance from 0 to A/2.

Step 4: Therefore, $T_1 < T_2$ .

Why other options are wrong:
- Option B: ${T_1} > {T_2}$ : This is incorrect because the particle is faster closer to the mean position.
- Option C: ${T_1} = {T_2}$ : This is incorrect because the speed is not constant.
- Option D: ${T_1} = 2{T_2}$ : This is incorrect because the speed decreases as it moves away from the mean position, so $T_2$ will be larger than $T_1$ , but not necessarily by a factor of 2.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Velocity in SHM

Memory Aids:
- Remember that velocity is maximum at the mean position and minimum at the extreme positions in SHM.

Quick Tips:
- Consider the variation of velocity with position in SHM to determine the time taken to travel different distances.

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Application +4 | -1

A particle undergoes SHM with a time period of 2 seconds. In how much time will it travel from its mean position to a displacement equal to half of its amplitude?

View answer and explanation

Correct option: D

\frac{1}{6}s

Option D (

\frac{1}{6}s

) is correct because the particle's displacement is half its amplitude at T/12, and since the period is 2 seconds, the time is 2/12 = 1/6 seconds.

Solution approach

Correct Answer + Core Reason: $\frac{1}{6}s$ . The time taken to travel from the mean position to half the amplitude is T/12, which is 1/6 seconds.

Step-by-Step Solution:

Step 1: The equation for SHM starting from the mean position is $x(t) = A \sin(\omega t)$ , where $A$ is the amplitude and $\omega$ is the angular frequency.

Step 2: We want to find the time $t$ when $x(t) = \frac{A}{2}$ . So, $\frac{A}{2} = A \sin(\omega t)$ .

Step 3: This simplifies to $\sin(\omega t) = \frac{1}{2}$ . The angle whose sine is $\frac{1}{2}$ is $\frac{\pi}{6}$ radians (30 degrees).

Step 4: Therefore, $\omega t = \frac{\pi}{6}$ .

Step 5: The angular frequency is related to the time period by $\omega = \frac{2\pi}{T}$ . Given $T = 2$ seconds, $\omega = \frac{2\pi}{2} = \pi$ rad/s.

Step 6: Substitute $\omega = \pi$ into $\omega t = \frac{\pi}{6}$ : $\pi t = \frac{\pi}{6}$ .

Step 7: Solve for $t$ : $t = \frac{\pi}{6\pi} = \frac{1}{6}$ seconds.

Why other options are wrong:
- **Option A ( $\frac{1}{2}s$ ): This is incorrect. It would be the time to reach the full amplitude if the motion was linear, but it's SHM.
- Option B ( $\frac{1}{3}s$ ): This is incorrect. It doesn't correspond to any significant point in the SHM.
- Option C ( $\frac{1}{4}s$ ): This is incorrect. It doesn't correspond to any significant point in the SHM.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Equation of SHM
- Angular frequency
- Time period

Memory Aids:**
- $x(t) = A \sin(\omega t)$ (SHM starting from mean position)
- $\omega = \frac{2\pi}{T}$

Quick Tips:
- Remember the standard angles and their sine values. $\sin(30^\circ) = \sin(\frac{\pi}{6}) = \frac{1}{2}$ .
- Recognize that the time to reach half the amplitude is not simply half the time to reach the full amplitude in SHM.

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Application +4 | -1

A conical pendulum of length 1 m makes an angle

\theta = {45^0}

w.r.t. Z-axis and moves in a circle in the XY plane. The radius of the circle is 0.4 m. and its centre is vertically below O. The speed of the pendulum, in its circular path, will be

\left( {g = 10\mathrm{ }m{s^{ - 2}}} \right)

View answer and explanation

Correct option: D

2 m/s

Option D (2 m/s) is correct because it's calculated using the radius of the circular path and the angle with respect to the vertical axis.

Solution approach

Correct Answer + Core Reason: 2 m/s. The speed is calculated from the radius, angle, and gravity.

Step-by-Step Solution:

Step 1: Given: length $l = 1 \mathrm{~m}$ , angle $\theta = 45^0$ , radius $r = 0.4 \mathrm{~m}$ , $g = 10 \mathrm{~m/s^2}$ .

Step 2: The radius of the circular path is related to the length and angle by $r = l \sin{\theta}$ .

Step 3: The speed of the pendulum in its circular path is given by $v = \sqrt{gr \tan{\theta}}$ .

Step 4: Substitute the given values: $v = \sqrt{10 \times 0.4 \times \tan{45^0}} = \sqrt{10 \times 0.4 \times 1} = \sqrt{4} = 2 \mathrm{~m/s}$

Why other options are wrong:
- Option A: Incorrect. This value doesn't match the correct calculation using the given parameters.
- Option B: Incorrect. This value doesn't match the correct calculation using the given parameters.
- Option C: Incorrect. This value doesn't match the correct calculation using the given parameters.

Key Concepts:
- Conical Pendulum
- Relationship between speed, radius, and angle in a conical pendulum

Memory Aids:
- Remember the formula for the speed of a conical pendulum: $v = \sqrt{gr \tan{\theta}}$ .

Quick Tips:
- Make sure to use the correct units for all the given values.
- Remember that $\tan{45^0} = 1$ .

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Remembering +4 | -1

The Q factor for a damped oscillator is given
by the

View answer and explanation

Correct option: C

Ratio of energy stored per cycle to the energy dissipated per cycle

Option C (Ratio of energy stored per cycle to the energy dissipated per cycle) is correct because the Q factor represents the sharpness of resonance, which is related to the ratio of stored energy to dissipated energy.

Solution approach

Correct Answer + Core Reason: Ratio of energy stored per cycle to the energy dissipated per cycle. The Q factor represents the sharpness of resonance.

Step-by-Step Solution:

Step 1: The Q factor (Quality factor) is a dimensionless parameter that describes how underdamped an oscillator or resonator is.

Step 2: It is defined as the ratio of the energy stored in the oscillator to the energy dissipated per cycle:
$Q = 2\pi \times \frac{\text{Energy stored}}{\text{Energy dissipated per cycle}}$
Or approximately, $Q = \frac{\text{Energy stored per cycle}}{\text{Energy dissipated per cycle}}$

Step 3: A high Q factor indicates a low rate of energy dissipation relative to the stored energy, meaning the oscillator will oscillate for a longer time.

Why other options are wrong:
- Option A: Ratio of energy stored per cycle to the initial energy is incorrect because the Q factor relates stored energy to dissipated energy, not initial energy.
- Option B: Ratio of energy dissipated per cycle to the initial energy is incorrect because the Q factor relates stored energy to dissipated energy, not initial energy.
- Option D: Ratio of energy dissipated per cycle to the energy stored per cycle is incorrect because the Q factor is the ratio of stored energy to dissipated energy.

Key Concepts:
- Q factor (Quality factor)
- Damped oscillator

Memory Aids:
- Remember that Q factor is related to the energy stored and dissipated in an oscillator.

Quick Tips:
- A high Q factor means less damping and longer oscillations.

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Application +4 | -1

Two particles P and Q start from origin and execute Simple Harmonic Motion along X-axis with same amplitude but with period 3 seconds and 6 seconds respectively. The ratio of the velocities of P and Q when they meet is

View answer and explanation

Correct option: B

\;2\mathrm{ }:1

Option B,

\;2\mathrm{ }:1

, is correct because the ratio of velocities is inversely proportional to the ratio of their time periods when they meet at the same displacement.

Solution approach

Correct Answer + Core Reason: $\;2\mathrm{ }:1$ . The ratio of velocities is inversely proportional to the ratio of their time periods.

Step-by-Step Solution:

Step 1: Write the equation for displacement in SHM: $x = A\sin(\omega t)$ , where $A$ is amplitude and $\omega$ is angular frequency.

Step 2: The particles meet when their displacements are equal: ${x_P} = {x_Q}$ . Therefore, ${A\sin({\omega_P}t)} = {A\sin({\omega_Q}t)}$ . This implies ${\omega_P}t = {\omega_Q}t + n\pi$ or ${\omega_P}t = \pi - {\omega_Q}t + n\pi$ , where n is an integer.

Step 3: Since they start from the origin, they first meet when ${\omega_P}t = \pi - {\omega_Q}t \Rightarrow ({\omega_P} + {\omega_Q})t = \pi$ .

Step 4: Express angular frequency in terms of time period: $\omega = \frac{{2\pi}}{T}$ . Therefore, $(\frac{{2\pi}}{{{T_P}}} + \frac{{2\pi}}{{{T_Q}}})t = \pi \Rightarrow (\frac{1}{{{T_P}}} + \frac{1}{{{T_Q}}})t = \frac{1}{2}$ .

Step 5: Given ${T_P} = 3$ s and ${T_Q} = 6$ s, substitute these values into the equation: $(\frac{1}{3} + \frac{1}{6})t = \frac{1}{2} \Rightarrow (\frac{1}{2})t = \frac{1}{2} \Rightarrow t = 1$ s.

Step 6: Write the equation for velocity in SHM: $v = A\omega\cos(\omega t)$ .

Step 7: Find the ratio of velocities: $\frac{{{v_P}}}{{{v_Q}}} = \frac{{{A{\omega_P}\cos({\omega_P}t)}}}{{{A{\omega_Q}\cos({\omega_Q}t)}}} = \frac{{{\omega_P}\cos({\omega_P}t)}}{{{\omega_Q}\cos({\omega_Q}t)}}$ .

Step 8: Substitute ${\omega_P} = \frac{{2\pi}}{3}$ and ${\omega_Q} = \frac{{2\pi}}{6} = \frac{\pi}{3}$ , and $t = 1$ s: $\frac{{{v_P}}}{{{v_Q}}} = \frac{{\frac{{2\pi}}{3}\cos(\frac{{2\pi}}{3})}}{{\frac{\pi}{3}\cos(\frac{\pi}{3})}} = \frac{{2\cos(\frac{{2\pi}}{3})}}{{\cos(\frac{\pi}{3})}} = \frac{{2(-\frac{1}{2})}}{{\frac{1}{2}}} = -2$ . Since we are looking for the ratio of magnitudes, the ratio is 2:1.

Why other options are wrong:
- Option A: $1\mathrm{ }:2$ This is the inverse of the correct ratio. It might be obtained by incorrectly assuming the velocities are directly proportional to the time periods.
- Option C: $2\mathrm{ }:3$ This ratio is incorrect and doesn't follow from the correct application of SHM equations and the given time periods.
- Option D: $3\mathrm{ }:2$ This ratio is also incorrect and doesn't follow from the correct application of SHM equations and the given time periods.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Angular Frequency
- Velocity in SHM

Memory Aids:
- $x = A\sin(\omega t)$ (Displacement in SHM)
- $v = A\omega\cos(\omega t)$ (Velocity in SHM)
- $\omega = \frac{{2\pi}}{T}$ (Angular frequency and time period)

Quick Tips:
- When dealing with SHM problems, always start by writing the equations for displacement and velocity.
- Pay attention to the initial conditions and use them to determine the phase constant.

29 Medium

29 of 94Question permalink

Application +4 | -1

Two mutually perpendicular simple harmonic vibrations have same amplitude, frequency and phase. When they superimpose, the resultant from of vibration will be

View answer and explanation

Correct option: C

A straight line

Option C (A straight line) is correct because when two SHMs with the same amplitude, frequency, and phase are perpendicular, their superposition results in a straight line with a slope of 1.

Solution approach

Correct Answer + Core Reason: A straight line. The superposition of two SHMs with the same amplitude, frequency, and phase, acting perpendicularly, results in a straight line.

Step-by-Step Solution:

Step 1: Let the two SHMs be $x = A \sin(\omega t)$ and $y = A \sin(\omega t)$ .

Step 2: Since both SHMs have the same amplitude, frequency, and phase, we can write $y = x$ .

Step 3: This equation represents a straight line passing through the origin with a slope of 1.

Why other options are wrong:
- Option A (A circle): This is incorrect because a circle would require a phase difference of $\frac{\pi}{2}$ between the two SHMs.
- Option B (An ellipse): This is incorrect because an ellipse would require different amplitudes or a phase difference other than 0 or $\frac{\pi}{2}$ .
- Option D (A parabola): This is incorrect because a parabola would require a different relationship between x and y, not a linear one.

Key Concepts:
- Superposition of SHM
- Lissajous Figures

Memory Aids:
- Remember that the shape of the resulting vibration depends on the amplitudes, frequencies, and phase difference of the individual SHMs.

Quick Tips:
- If the two SHMs have the same frequency and a phase difference of 0 or $\pi$ , the resulting vibration will be a straight line.
- If the two SHMs have the same frequency and a phase difference of $\frac{\pi}{2}$ and the same amplitude, the resulting vibration will be a circle.

30 Medium

30 of 94Question permalink

Application +4 | -1

The minimum phase difference between two simple harmonic oscillations,

{y_1} = \frac{1}{2}\sin \,\,\omega t + \frac{{\sqrt 3 }}{2}\cos \,\,\omega t

{y_2} = \sin \,\,\omega t + \cos \,\,\omega t,

is

View answer and explanation

Correct option: B

\frac{\mathrm{\pi }}{{12}}

Option B (

\frac{\mathrm{\pi }}{{12}}

) is correct because it represents the minimum phase difference calculated after converting the given equations to standard sine form.

Solution approach

Correct Answer + Core Reason: $\frac{\mathrm{\pi }}{{12}}$ . The phase difference is calculated by converting both SHM equations to standard sine form and then finding the difference in their phase angles.

Step-by-Step Solution:

Step 1: Rewrite the first equation: ${y_1} = \frac{1}{2}\sin \omega t + \frac{{\sqrt 3 }}{2}\cos \omega t$ . This can be written as ${y_1} = \sin(\omega t + \phi_1)$ , where $\phi_1$ is such that $\cos \phi_1 = \frac{1}{2}$ and $\sin \phi_1 = \frac{{\sqrt 3 }}{2}$ . Therefore, $\phi_1 = \frac{\pi}{3}$ . So, ${y_1} = \sin(\omega t + \frac{\pi}{3})$ .

Step 2: Rewrite the second equation: ${y_2} = \sin \omega t + \cos \omega t$ . This can be written as ${y_2} = A\sin(\omega t + \phi_2)$ , where $A = \sqrt{1^2 + 1^2} = \sqrt{2}$ . Thus, ${y_2} = \sqrt{2} \sin(\omega t + \phi_2)$ , where $\cos \phi_2 = \frac{1}{\sqrt{2}}$ and $\sin \phi_2 = \frac{1}{\sqrt{2}}$ . Therefore, $\phi_2 = \frac{\pi}{4}$ . So, ${y_2} = \sqrt{2}\sin(\omega t + \frac{\pi}{4})$ .

Step 3: Find the phase difference: $\Delta \phi = |\phi_1 - \phi_2| = |\frac{\pi}{3} - \frac{\pi}{4}| = |\frac{{4\pi - 3\pi}}{{12}}| = \frac{\pi}{12}$ .

Final Answer: $\frac{\pi}{12}$

Why other options are wrong:
- **Option A ( $\frac{{7\pi }}{{12}}$ ): This is incorrect because it results from an incorrect calculation of the phase difference or incorrect identification of the phase angles.
- Option C ( $- \frac{\pi }{6}$ ): This is incorrect because phase difference is usually taken as a positive value or the absolute difference, and the calculation is wrong.
- Option D ( $\frac{\pi }{6}$ ): This is incorrect because it results from an incorrect calculation of the phase difference or incorrect identification of the phase angles.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Phase difference
- Trigonometric identities
- Superposition of SHMs

Memory Aids:**
- Remember the formula for combining sine and cosine terms: $a\sin \theta + b\cos \theta = \sqrt{a^2 + b^2} \sin(\theta + \phi)$ , where $\phi = \arctan(\frac{b}{a})$ .

Quick Tips:
- Always convert the SHM equations to standard sine or cosine form before calculating the phase difference.
- Be careful with trigonometric identities and phase angle calculations.

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